Question

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant $100N{m}^{-1.}$ A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

Answer 1

Amplitude = 0.1 m

Total mass = $3+1=4kg$

(when both the blocks are moving together)

$K=100N/M$

$\therefore T=2\pi \frac{m}{k}$

$=2\pi \sqrt{\frac{4}{100}}=\frac{2\pi }{5}sec$

$\therefore$ Frequency = $\frac{5}{2\pi }Hz$

Again at the mean position, let 1 kg block has velocity v.

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

where $x\to$ Amplitude = 0.1 m

$\therefore \left(\frac{1}{2}\right)\times (1\times v^2)(1\times 2)\times 100(0.1{)}^2$

$Kv=1mse{c}^{-1}...\left(1\right)$

After the 3 kg block is gently placed on the 1 kg, then let, $1kg+3kg=4kg$ block and the spring the one system. For this mass spring system there is no external force. (When oscillation takes place). The momentum should be conserved. Let 4 kg block has velocity v.

$\therefore$ Initial momentum = Final momentum

$\therefore 1\times v=4\times v^1$

$\Rightarrow v^1=\frac{1}{4}m/s$

[as $v=1mse{c}^{-1}$ from equation (i)]

Now the two blocks have velocity $\frac{1}{4}m{s}^{-1}$ at its mean position.

KE mean = $\frac{1}{2}m{v}^2$

$=\left(\frac{1}{2}\right)\times 4\times {\left(\frac{1}{4}\right)}^2=\frac{1}{2}\times \frac{1}{4}$

KE mean $=\left(\frac{1}{2}\right)m^2{v}^2\left(\frac{1}{2}\right)4x={\left(\frac{1}{4}\right)}^2$

$=\frac{1}{2}\times \frac{1}{4}$

When the blocks are going to the extrem position, there will be only potential energy

$\therefore PE=\frac{1}{2}k{\delta }^2=\frac{1}{2}\times \frac{1}{4}$

where $d\to$ New amplitude

$\therefore \frac{1}{4}=100{\delta }^2$

$=d=\sqrt{\left(\frac{1}{400}\right)}$

$=0.05m=5cm$

So, amplitude = 5 cm